erl - 如何检查一个元素是否包含在列表中？

Question

如何从列表中捕获单词并True在 Erlang 中返回？

catch_word(msg) ->
    Bad = ["BadWord1", "BadWord2"],
    case Bad in msg of
        true ->
            true;
        false ->
            false
    end.

catch_word("Hello, How are u BadWord1").

Answer 1

欢迎来到 Erlang，你可能想试试这个：

-export([catch_word/2]).

catch_word(Msg,BadWords)->
    catch_word(Msg,BadWords,0).

catch_word(Msg,[],0)->
    false;
catch_word(Msg,[Word|BadWords],0)->
        catch_word(Msg,BadWords,string:str(Msg,Word));
catch_word(_,_,N)->
    true.

和

your_module:catch_word("Hello, How are u BadWord1",["BadWord1", "BadWord2"]).